Put a section of round steel with a radius of 5 cm into the water, and the water level will rise by 9 cm.
This shows that the volume of the whole round steel is the same as that when the water level of the bucket rises by 9 cm.
You can get the equation.
5*5*π*h=r*r*π*9
That is, there is an equation-25h = 9R2 (the square of R).
After the round steel is vertically pulled out of the water by 8 cm, the water level drops by 4 cm.
This shows that the volume of round steel immersed in water at this time is
5*5*π*(h-8)
At this time, the water level of the bucket rose by 9-4=5 cm.
free
5*5*π*(h-8)=5*π*r*r
That is, Equation 2, 25h-200 = 5R2.
One and two simultaneous cubic equations can be solved.
H= 18 (cm)
At this time, the volume of round steel is
V = 5 * 5 *18 * π = 450 π ≈1413 (cubic centimeter)